Example 1

that passes through the point (0,0). Use Euler's Method to estimate y(1) with step size Δ x = 0.5.

We want to take steps of size 0.5 to get from x = 0 to x = 1, so there will be two steps.

This means we'll do two tangent line approximations. First we'll draw a tangent line to y at 0 and use that to estimate y(0.5).

Then we'll draw a line at x = 0.5 and use that to estimate y(1).

First Tangent Line Approximation: We're going to use a tangent line to y at x = 0 to estimate y(.5), so Δ x = .5. Since y(0) = 0 we have

yold = 0.

The slope of the tangent line at (0,0) is

Our tangent line is horizontal.

We calculate

We estimate

y(.5) ≅ 0.

Second Tangent Line Approximation: The idea is to use a tangent line to y at x = 0.5 to estimate y(1). However, we only have an estimate of y(0.5). We don't know its real value. This means the line we draw won't really be tangent to y at x = 0.5, but it will be pretty close.

We have the point (0.5, 0) to start from, so we let yold = 0. We still have Δ x = 0.5. To get the slope of the line we use the differential equation:

Now we can estimate y(1):

Our final estimate, using Euler's Method, is

y(1) ≅ 0.5.

Example 2

Let y be the solution to the differential equation

that passes through the point (0,0). Use Euler's Method to estimate y(1) with step size Δ x = .2.

Knowing the step size means we know each value of x we'll need to look at, so we can put these values in the table:

x

yold

slope × Δ x

ynew = yold + slope × Δ x

0

0.4

0.4

0.6

0.8

1

We know the initial value y(0) = 0, so we can put that in:

Now we start the tangent line approximations.

From the point (0,0) we calculate

so we can fill in more of the table:

Now we take ynew and use it as the value of yold for the next step:

When x = 0.2 and y = 0 the slope is 2(0.2) = .4, so we can fill in more:

When x = 0.4 and y = 0.08 we get the next step:

When x = .6 and y = .24 we get this:

And finally, when x = .8 and y = .48 we get this:

Since we now have a y-value corresponding to x = 1, we're done.We conclude that for the given IVP, Euler's method with Δ x = .2 estimates y(1) ≅ .8

An Euler's Method problem must include

an initial value

enough information so you can figure out Δ x

the stopping point (that is, the value of x at which you want to approximate the function).

The previous example said

"Let y be the solution to the differential equation

that passes through the point (0,0). Use Euler's Method to estimate y(1) with step size Δ x = 0.2."

The initial value is the point (0,0). Δ x = 0.2 was explicitly given. Since the problem says to estimate y(1), we know x = 1 is the stopping point.

The problem could also have said "Use Euler's Method with 5 steps to estimate y(1)." In this case, the problem doesn't explicitly tell you Δ x. But it does tell you that you're supposed to get from x = 0 to x = 1 in 5 steps, which means

Example 3

The function y is a solution to the differential equation

with y(0) = 5. Estimate y(1) using Euler's method with two steps.

To get from 0 to 1 in two steps means each step must be .5, so Δ x = .5. Knowing Δ x and knowing that y(0) = 5, we can start the table:

When x = 0 and y = 5 we have

When x = .5 and y = 5 we have

We conclude that

y(1) ≅ 4.95

After the first step, we have only approximate values of y to look at. Since the formula for the derivative requires both x and y, the best we can do is use our approximate values of y.

This is what we meant earlier when we said we were using pretend tangent lines. Instead of a point on the function, we have a point close to the function. Instead of having the slope of the function at that point, we have the slope the function would have if it did pass through that point.